Study Notes for Current Density

Chapter IV: Current Density

1. Electric Current

1.1 Definition

  • Electric Current: Defined as any collective motion of charged particles.

  • Intensity of Electric Current (i): The amount of charge dqdq crossing a surface dSdS per unit time dtdt, mathematically represented as:   - i=racdqdti = rac{dq}{dt}

1.2 Volume Current Density j\boldsymbol{j}

  • Considerations of a volume-charged cylinder containing moving charge carriers (specifically electrons e−e^-).

  • Notation:   - dS\boldsymbol{dS}: The elementary surface.   - v\boldsymbol{v}: The velocity of the charge carriers.   - dVdV: The corresponding infinitesimal volume.

Elementary Volume Relation

  • The elementary volume is derived from the product of the surface dS\boldsymbol{dS} and its displacement dl\boldsymbol{dl}:   - dV=dS∙dldV = \boldsymbol{dS} \bullet \boldsymbol{dl}

  • Between time tt and t+dtt + dt, the average distance traveled by the carriers is given by:   - dl=vdt\boldsymbol{dl} = \boldsymbol{v} dt   - Therefore, the elementary volume can be rewritten as:   - dV=dS∙vdtdV = \boldsymbol{dS} \bullet \boldsymbol{v} dt

  • If the surface element dS\boldsymbol{dS} is collinear with the velocity of the charge carriers v\boldsymbol{v}, then,   - dV=dSimesvdtdV = dS imes v dt

Charge in Volume

  • The amount of charge dqdq in this volume dVdV is defined by:   - dq=ho(r)dV=ho(r)dS∙vdtdq = ho(r)dV = ho(r)dS \bullet v dt where ho(r)ho(r) is the volume charge density.

  • This density can be associated with the number of carriers nn, each with charge qq:   - ho(r)=nqho(r) = nq

  • Thus, substituting:   - dq=ho(r)dV=nqdSvdtdq = ho(r)dV = nqdSvdt

Current Intensity Expression

  • The expression for current intensity becomes:   - i=racdqdt=nqvdS=ho(r)vdS=ho(r)v∙dSi = rac{dq}{dt} = nqvdS = ho(r)vdS = ho(r) \boldsymbol{v} \bullet \boldsymbol{dS}

Volume Current Density Vector

  • For the distribution of moving charges with average velocity v\boldsymbol{v},   - Volume Current Density Vector: Denoted as j(r)\boldsymbol{j}(r):     - j(r)=ho(r)v\boldsymbol{j}(r) = ho(r) \boldsymbol{v}

  • The total current can then be defined as:   - I=intSj(r)∙dSI = int_{S} \boldsymbol{j}(r) \bullet \boldsymbol{dS}

  • Note that j\boldsymbol{j} represents volume current density, despite the integration over a surface dSdS.

1.3 Surface Current Density jS\boldsymbol{j}_S

  • When currents are confined near a surface SS with thickness approaching zero (eightarrow0e ightarrow 0):   - dS=dlimeseext(approx.equalto:dl)dS = d\boldsymbol{l} imes e ext{ (approx. equal to: } d\boldsymbol{l})

  • The total current then is represented as:   - I=intLjS∙dlI = int_{L} \boldsymbol{j}_S \bullet d\boldsymbol{l}

2. Application Exercises

Exercise 1: Conduction in a Copper Wire

  • Given parameters:   - Cross-sectional area of the wire: S=1.0extmm2S = 1.0 ext{ mm}^2   - Current through the wire: I=1.0extAI = 1.0 ext{ A}   - Conductivity of copper: extdenotedbyβext{denoted by } \beta   - Molar mass: M=6.35imes10−2extkgmol−1M = 6.35 imes 10^{-2} ext{ kg mol}^{-1}   - Mass density: ho=8.95imes103extkgm−3ho = 8.95 imes 10^{3} ext{ kg m}^{-3}   - Avogadro’s number: NA=6.02imes1023extmol−1N_A = 6.02 imes 10^{23} ext{ mol}^{-1}   - Each copper atom releases one conduction electron, charge: q=−e,extwithe=1.6imes10−19extCq = -e, ext{ with } e = 1.6 imes 10^{-19} ext{ C}

Solutions:

  1. Volume Density of Charge Carriers npn_p:      - Formula:np=rachoMNAn_p = rac{ ho}{M} N_A      - Calculation: np=rac8.95imes103extkgm−36.35imes10−2extkgmol−1imes6.02imes1023extmol−1ightarrownpextapprox.8.5imes1028extm−3n_p = rac{8.95 imes 10^3 ext{ kg m}^{-3}}{6.35 imes 10^{-2} ext{ kg mol}^{-1}} imes 6.02 imes 10^{23} ext{ mol}^{-1} ightarrow n_p ext{ approx. } 8.5 imes 10^{28} ext{ m}^{-3}   2. Volume Current Density:      - Formula: j=racISez\boldsymbol{j} = rac{I}{S} \boldsymbol{e_z}, where ez\boldsymbol{e_z} is the unit vector along wire axis.      - Therefore,        - Magnitude: j=racIS=rac1extA1imes10−6extm2=106extAm−2j = rac{I}{S} = rac{1 ext{ A}}{1 imes 10^{-6} ext{ m}^2} = 10^6 ext{ A m}^{-2}   3. Drift Velocity of Electrons:      - Using the relation: j=npqvd\boldsymbol{j} = n_p q \boldsymbol{v_d}:        - Rearranging gives total drift velocity:          - vd=racjnpqv_d = rac{j}{n_p q}          - Substituting values: vdextapprox.7.4imes10−5extms−1v_d ext{ approx. } 7.4 imes 10^{-5} ext{ m s}^{-1}      - Conclusion: The drift velocity is extremely small, on the order of tens of micrometers per second.

Exercise 2: Calculation of Electrical Resistance

  • Consider a cylindrical conductor with inner radius R1R_1 and outer radius R2R_2 (R_2 > R_1) with length l\boldsymbol{l} and conductivity β\beta:

Derivation of Resistance

  1. Set-Up:    - Elementary Cylindrical Shell: Radius rr and thickness drdr.    - Thus, the conduction surface is given as:      - S=2imesextπrlS = 2 imes \boldsymbol{ ext{π}} r \boldsymbol{l}

  2. Elementary Resistance:    - dR=racdrβS=racdrβ2extπrldR = rac{dr}{\beta S} = rac{dr}{\beta 2 \boldsymbol{ ext{π}} r \boldsymbol{l}}

  3. Total Resistance Calculation: Integrate between the limits R1R_1 and R2R_2,    - R=intR1R2dR=rac12extπβlintR1R2racdrrR = int_{R_1}^{R_2} dR = rac{1}{2 \boldsymbol{ ext{π}} \beta \boldsymbol{l}} int_{R_1}^{R_2} rac{dr}{r}
       - Final expression:      - R=rac12extπβlextln(racR2R1)R = rac{1}{2 \boldsymbol{ ext{π}} \beta \boldsymbol{l}} ext{ln} \bigg( rac{R_2}{R_1}\bigg)

Conclusion

  • The understanding of electric current density is essential in electrical and electronic engineering applications. Exercises illustrate practical applications in current flow and resistance calculations. The lesson integrates theoretical principles with real-world material properties, empowering students to derive insights about current and conductivity in practical scenarios.